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Volume, surface, and line integrals

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider a vector A = (x^2 - y^2)(i) + xyz(j) - (x + y + z)k and a cube bounded by the planes x = 0, x = 1, y = 0, y = 1, z = 0 and z = 1
    Determine the volume integral ∫∇.A dV where V is the volume of the cube
    Determine the surface integral ∫A.n dS where s is the surface of the cube



    2. Relevant equations



    3. The attempt at a solution
    ∇.A = 2x + xz -1
    Volume integral = ∫∫∫(all from 0 to 1) (2x + xz -1)dxdydz
    =1/4 (after simplifying)

    Surface integral = (all from 0 to 1) ∫∫(x^2-y^2)dydz + ∫∫(x+y+z)dxdy + ∫∫(xyz)dxdz
    this simplifies to a more complicated term

    I know that both of these methods must lead to the same answer, so I know that I must be doing something wrong with assigning the integrals to evaluate. Can someone show me how to properly set up the volume and surface integrals? This is what I'm confused about the most.
     
  2. jcsd
  3. Nov 18, 2013 #2

    vela

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    Here's an example for one of the faces. For the x=1 face, the outward normal is given by ##\hat{n} = \hat{i}##, so ##\vec{A}\cdot\hat{n} = x^2-y^2 = 1-y^2##. The flux through that face is therefore
    $$\int_0^1 \int_0^1 (1-y^2)\,dy\,dz = \frac{2}{3}.$$ Try working out the rest on your own and show us what you get.
     
  4. Nov 18, 2013 #3
    So one of the variables will be a constant for each side evaluated? So for,
    integral of (x+y+z)dxdy
    =(1+x+y)dxdy
    =(x + x^2/2 + xy)dy
    = (1 + 1/2 + y)dy
    =y + 0.5y + y^2/2
    = 1 + 0.5 + 0.5
    =2

    and integral of (xyz)dxdz from 0 to 1
    =(1xz)dxdz
    =(x^2/2)*z(dz)
    =0.5z(dz)
    =0.5z^2/2
    =0.25

    so 2 + 0.25 + 2/3 =35/12

    Does this mean my volume integral is wrong?
     
  5. Nov 18, 2013 #4

    vela

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    You have to calculate the flux for all six faces of the cube. You've only done half of them so far.
     
  6. Nov 18, 2013 #5
    So I use the same integrals listed above but compute them for the x = 0, y = 0, and z = 0 face, then add all the integrals together?
    So the first x = 0 face it would be
    -(-y^2)dydz
    (y^3/3)dz
    1/3

    z = 0
    -(x + y)dxdy
    -(x^2/2 + xy)dy
    -(0.5 + y)dy
    -1.5

    y = 0
    0

    so 1/3 - 1.5 + 2/3 +2 + 0.25
    = 1.75

    Edit: nevermind, I'm doing something completely wrong. I just calculate the opposite faces, but wouldn't they lead to the negatives of the above?
     
    Last edited: Nov 18, 2013
  7. Nov 18, 2013 #6

    vela

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    No, they won't simply be the negatives. Recheck your calculations for the z=0 and z=1 faces.
     
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